# Deriving the Formula of the Vertex of Quadratic Functions

In getting the vertex of the quadratic function in general form $f(x) = ax^2 + bx + c$, we usually need to convert it to the vertex form $f(x) = a(x - h)^2 + k$. In the latter form, the vertex of the parabola is at $(h,k)$. For example, the function in the general form $f(x) = 2x^2 - 12x + 22$

can be rewritten in the vertex form as $f(x) = 2(x - 3)^2 + 4$.

In the vertex form, it is easy to see that the vertex is at $(h,k) = (3,4)$.

Aside from this method, we can also use the ordered pair $\left ( \dfrac{-b}{2a}, \dfrac{4ac - b^2}{4a} \right )$

in place of $(h,k)$

This means that we can get the vertex of $f$ without changing the general form to vertex from. In this post, we are going to derive the ordered pair above. That is, we are going to show that $h = \dfrac{-b}{2a}$

and $k = \dfrac{4ac - b^2}{4a}$.

Expanding $f(x) = a(x - h)^2 + k$ gives us $f(x) = a(x^2 - 2hx + h^2) + k$ $f(x) = ax^2 - 2ahx + ah^2 + k$.

Now, the terms with both x’s in the general and vertex forms are $bx$ and $-2ahx$. So, we can equate the two expressions and solve for $h$. $-2ahx = bx$ $-2ah = b$ $h = \frac{- b}{2a}$.

Now, we can use $h$ as the value of $x$ in the general from. That is $f(h) = ah^2 + bh + c$ $f\left ( \dfrac{-b}{2a} \right ) = a \left ( \dfrac{-b}{2a} \right )^2 + b \left (\dfrac{-b}{2a} \right ) + c$ $= a \left (\dfrac{b^2}{4a^2} \right ) - \dfrac{-b^2}{2a} + c$ $= \dfrac{b^2}{4a} - \dfrac{b^2}{2a} + c$ $= \dfrac{b^2 - 2b^2 + 4ac}{4a}$ $= \dfrac{4ac - b^2}{4a}$ $k = \dfrac{4ac - b^2}{4a}$ .

This means that we can get the vertex of the function $f(x) = 2x^2 - 12x + 22$

without converting it to the vertex form as follows. $h = \dfrac{-b}{2a} = \dfrac{-(-12)}{2(2)} = 3$ $k = \dfrac{4(2)(22) - (-12)^2}{4(2)} = \dfrac{32}{8} = 4$

So, the vertex is at $(h,k) = (3,4)$ which is consistent with our calculation above.